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CP Stopping Distance. (a) If the coefficient of kinetic
Chapter 8, Problem 33E(choose chapter or problem)
CP Stopping Distance.? (a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (?Note:? Locking the brakes is ?not? the safest way to stop.)
Questions & Answers
QUESTION:
CP Stopping Distance.? (a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (?Note:? Locking the brakes is ?not? the safest way to stop.)
ANSWER:Solution 33E After locking the brake, the car will slide on the ground, and because of the kinetic friction there will be deceleration and using the deceleration we can find out the distance required to stop the car. In the second part, we know the distance and we can calculate the deceleration of the car from dynamic friction. Using this two we can calculate the maximum speed. Step 1 Let us consider the mass of the car is m, hence the force due to kinetic friction is F = mg kf From the problem we know that, = 0.80 We also know that g = 9.8 m/s 2 Hence the force due to kinetic friction is 2 F kf = m(0.80)(9.8 m/s ) Fkf m(0.80)(9.8 m/s ) 2 So the deceleration of the car is a = m = g = m = 7.84 m/s